567.Permutation-in-String

567. Permutation in String

题目地址

https://leetcode.com/problems/permutation-in-string/

题目描述

Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.

Example 1:
Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:
Input:s1= "ab" s2 = "eidboaoo"
Output: False

Note:
The input strings only contain lower case letters.
The length of both given strings is in range [1, 10,000].

代码

Approach #1 Using Array Map

class Solution {
  public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length())    return false;
    int[] s1map = new int[26];
    for (int i = 0; i < s1.length(); i++) 
      s1map[s1.charAt(i) - 'a']++;
    for (int i = 0; i < s2.length() - s1.length(); i++) {
      int[] s2map = new int[26];
      for (int j = 0; j < s1.length(); j++) {
        s2map[s2.charAt(i + j) - 'a']++;
      }
      if (matches(s1map, s2map)) {
        return true;
      }
    }
    return false;
  }

  private boolean matches(int[] s1map, int[] s2map) {
    for (int i = 0; i < 26; i++) {
      if (s1map[i] != s2map[i]) {
        return false;
      }
    }
    return true;
  }
}

Approach #2 Sliding Windwow

public class Solution {
    public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length())
            return false;
        int[] s1map = new int[26];
        int[] s2map = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            s1map[s1.charAt(i) - 'a']++;
            s2map[s2.charAt(i) - 'a']++;
        }
        for (int i = 0; i < s2.length() - s1.length(); i++) {
            if (matches(s1map, s2map))   return true;

            s2map[s2.charAt(i + s1.length()) - 'a']++;
            s2map[s2.charAt(i) - 'a']--;
        }
        return matches(s1map, s2map);
    }
    public boolean matches(int[] s1map, int[] s2map) {
        for (int i = 0; i < 26; i++) {
            if (s1map[i] != s2map[i])
                return false;
        }
        return true;
    }
}

Approach #3 Optimized Sliding Window

public class Solution {
    public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length())
            return false;
        int[] s1map = new int[26];
        int[] s2map = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            s1map[s1.charAt(i) - 'a']++;
            s2map[s2.charAt(i) - 'a']++;
        }
        int count = 0;
        for (int i = 0; i < 26; i++)
            if (s1map[i] == s2map[i])
                count++;
        for (int i = 0; i < s2.length() - s1.length(); i++) {
            int r = s2.charAt(i + s1.length()) - 'a', l = s2.charAt(i) - 'a';
            if (count == 26)
                return true;
            s2map[r]++;
            if (s2map[r] == s1map[r])
                count++;
            else if (s2map[r] == s1map[r] + 1)
                count--;
            s2map[l]--;
            if (s2map[l] == s1map[l])
                count++;
            else if (s2map[l] == s1map[l] - 1)
                count--;
        }
        return count == 26;
    }
}